Article: 22715 of sci.math
From: cet1@cus.cam.ac.uk (Chris Thompson)
Subject: Rep-1 squares question
Summary: It's more difficult than it looks
Date: 15 Mar 93 00:51:27 GMT
Organization: U of Cambridge, England
David Wilson writes:
>
> 11111 is a square in base 3.
>
> Are there longer rep-1 (or rep-k) squares in other bases?
>
No-one seems to have followed up on this posting: I wonder why, as
it is not devoid of interest. Sticking to the rep-1 squares (quite
difficult enough) we are looking for non-negative integers x,y,n
satisfying
y^2 = x^n + ... + x + 1 = (x^(n+1)-1)/(x-1) = P_n(x), say
n=0 or x=0 always give solutions, n=1 or x=1 give many uninteresting
ones, so assume n>=2 and x>=2. A bit of experimenting will probably
suggest to anyone that the only solutions in this range are 20^2=P_3(7)
and 11^2=P_4(3). Is there are any hope of proving this?
It is easy to see that n=2 gives no solutions, as x^2+x+1 lies strictly
between x^2 and (x+1)^2 and therefore cannot be a square. One can use a
similar line of attack for any even n. Thus
(x^2+x/2)^2 < P_4(x) < (x~2+x/2+1)^2
showing that one cannot have n=4 and x even; while
(x^2+x/2+1/2)^2 - P_4(x) = (x+1)(x-3)/4
exhibiting the only solution with n=4, namely x=3,y=11. Continuing,
one can show that
P_6(x) - (x^3+x^2/2+3x/8+M/8)^2 = 0
has no (suitable) integer roots for each of M=0,1,2,3,4,5,6,7; hence
there are no solutions with n=6. Any particular even value of n can be
dealt with in this way, but can one deal with them all at once?
For n=3, y^2=x^3+x^2+x+1 is the equation of an elliptic curve in
canonical form; it is convenient to transform with X=x+1,Y=y to
Y^2=X(X^2-2X+2). Standard theory (see, for example, [1]) enables
one to show that the *rational* points on the curve are generated
by (X,Y) = (0,0), of order 2, and (1,1), of infinite order:
... inf (1,1) (1/4,-5/8) (49/9,-287/7) (961/400,21359/8000) ...
... (0,0) (2,-2) (8,20) (18/49,246/343) (800/961,-27560/29791) ...
But are there any *integer* solutions after (8,20) ? (i.e. x=7,y=20)
>From Thue-Siegel theory there are only finitely many; after Baker
(e.g. [2] Theorem 4.2) we know that they can in principle be
effectively determined. But can this be carried through in practice?
A more elementary approach would be to observe that y^2=(x+1)(x^2+1)
and gcd(x+1,x^2+1) | 2 means that we must have x+1 = 2a^2, x^2+1 = 2b^2.
The general solution to the latter is
x + b sqrt(2) = (1 + sqrt(2))^m with m odd
So we require (1 + sqrt(2))^m + (1 - sqrt(2))^m + 2 to be a square.
m=3 gives x=7,y=20. Is there anything to be got out of this approach?
Of course, y^2=x^4+x^3+x^2+x+1 is also an elliptic curve. The birational
tranformation
t = 2y + 2x^2 + x + 2
u = y(4x+1) + 4x^3 + 3x^2 + 2x + 1
converts it to the canonical form u^2=t(t^2-5t+5). As in the n=3 case
the rational points are generated by (t,u) = (0,0) of order 2 and (1,1)
of infinite order:
... (1/9,19/27) (4,-2) (1,-1) inf (1,1) (4,2) (1/9,-19/27) ...
... (45,-285) (5/4,5/8) (5,5) (0,0) (5,-5) (5/4,-5/8) (45,285) ...
corresponding to (x,y) values:
... (-8/11,101/121) (3,-11) inf (-1,-1) (0,1) (1/3,-11/9) ...
... (-123/35,*) inf (1/3,11/9) (0,-1) (-1,1) inf (3,11) ...
but in this case we know that we have all the integer points already
by the earlier argument.
For n>=5, y^2=P_n(x) is a curve of genus greater than 1 and will have
only finitely many *rational* points (Faltings), let alone integer ones.
This doesn't necessarily help in finding them, of course...
Can anyone finish off at least one of the open ends here?
[1] J.W.S.Cassels, "Lectures on Elliptic Curves",
Cambridge University Press 1991, ISBN 0-521-41517-9
[2] Alan Baker, "Trancendental Number Theory",
Cambridge Univeristy Press 1975, ISBN 0-521-20461-5
Chris Thompson
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